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vekshin1
3 years ago
10

Michele correctly solved a quadratic equation using the quadratic formula as shown below.

Mathematics
1 answer:
ra1l [238]3 years ago
8 0
Whats the answers it doesn’t show for me
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Answer:

In the graph of a circle, we can easily draw a straight vertical line passing through its center. This will hit the top and the bottom of the shape. Since this line hits two points, a circle's graph is not a function. We can also observe the equation of a circle.

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This should help you

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Which set of lengths are not the side lengths of a right triangle? (A-28, 45, 53) (B-13, 84, 85) (C-36, 77, 85) (D-16, 61, 65) [
nikklg [1K]
Side lengths 16, 61, and 65 are not part of a right triangle.

A right triangle's sides should follow this formula, C being the largest number:

a^{2} + b ^{2} = c^{2}

16^{2} = 256

61^{2} = 3721

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3 years ago
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jek_recluse [69]
I believe its B and D
6 0
3 years ago
Find the measure of angle eight in each triangle round each answer to the nearest 10th
vlada-n [284]

Answer:

# m∠A = 65.3°

# m∠A = 25.8°

# m∠A = 22.7°

Step-by-step explanation:

* Lets revise the trigonometry function to solve the problem

- In any right angle triangle:

# The side opposite to the right angle is called the hypotenuse

# The other two sides are called the legs of the right angle

* If the name of the triangle is ABC, where B is the right angle

∴ The hypotenuse is AC

∴ AB and BC are the legs of the right angle

- ∠A and ∠C are two acute angles

- For angle A

# sin(A) = opposite/hypotenuse

∵ The opposite to ∠A is BC

∵ The hypotenuse is AC

∴ sin(A) = BC/AC

# cos(A) = adjacent/hypotenuse

∵ The adjacent to ∠A is AB

∵ The hypotenuse is AC

∴ cos(A) = AB/AC  

# tan(A) = opposite/adjacent

∵ The opposite to ∠A is BC

∵ The adjacent to ∠A is AB

∴ tan(A) = BC/AB

* Lets solve the problems

# In Δ ABC

∵ m∠B = 90°

∵ AB = 2.3 ⇒ adjacent to angle A

∵ BC = 5 ⇒ apposite to angle A

- To find m∠A use the tangent function because we have opposite

  and adjacent sides

∴ tan A = BC/AB

∴ tan A = 5/2.3 ⇒ use tan^-1 to find m∠A

∴ m∠A = tan^{-1}\frac{5}{2.3}=65.29756

* m∠A = 65.3°

# In Δ ABD

∵ m∠B = 90°

∵ AB = 5.4 ⇒ adjacent to angle A

∵ DA = 6 ⇒ the hypotenuse

- To find m∠A use the cosine function because we have adjacent

  and hypotenuse sides

∴ cos A = AB/DA

∴ cos A = 5.4/6 ⇒ use cos^-1 to find m∠A

∴ m∠A = cos^{-1}\frac{5.4}{6}=25.84193

* m∠A = 25.8°

# In Δ ABE

∵ m∠B = 90°

∵ EB = 2.4 ⇒ opposite to angle A

∵ EA = 6.8 ⇒ the hypotenuse

- To find m∠A use the sine function because we have opposite

  and hypotenuse sides

∴ sin A = EB/EA

∴ sin A = 2.4/6.8 ⇒ use sin^-1 to find m∠A

∴ m∠A = sin^{-1}\frac{2.4}{6.8}=22.6673

* m∠A = 22.7°

3 0
3 years ago
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